# How do you find two consecutive odd integers whose sum is 196?

Apr 25, 2016

$97 \mathmr{and} 99$

#### Explanation:

There is at least two ways of doing this: This is my approach

$\textcolor{b l u e}{\text{Step 1}}$ Define a number so that it is always even

$\text{ }$Let $n$ be any number then $2 n$ is always even.

$\textcolor{b l u e}{\text{Step 2}}$ Modify the defined number so that the result is always odd

$\text{ }$If $2 n$ is always even then $2 n + 1$ is always odd
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Let the first odd number be $2 n + 1$
Then the second odd number is $\left(2 n + 1\right) + 2 = 2 n + 3$

Thus our given condition is such that:

$\left(2 n + 1\right) + \left(2 n + 3\right) = 196$

$\implies 4 n + 4 = 196$

Subtract 4 from both sides

$\implies 4 n = 192$

Divide both sides by 4

$\implies n = \frac{192}{4} = 48$

But the first number is $2 n + 1 = 2 \left(48\right) + 1 = 97$

So the second number is $97 + 2 = 99$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: $97 + 99 = 196$

Apr 25, 2016

Another way

#### Explanation:

Let the first number be $n$
Let the second number be $n + 2$

Then $n + n + 2 = 196$

$2 n + 2 = 196$

Subtract 2 from both sides

$2 n = 194$

Divide both sides by 2

$n = \frac{194}{2} = 97$

The first number is 97 so the second is 97+2=99

$\textcolor{red}{\text{Notice that the " 2n + 2" is of the same format as that in my}}$$\textcolor{red}{\text{other solution of } 4 n + 4}$